Question

Consider the following data for a substance: Critical point: 114.9 atm and 555°C Triple point: 0.103...

Consider the following data for a substance: Critical point: 114.9 atm and 555°C Triple point: 0.103 atm and 114°C Normal melting point: 1.00 atm and 125°C What is the normal boiling point of the substance? Enter your answer in units of °C to three significant figures.

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Answer #1

The boiling point, triple point and critical point are the three points in the vaporization curve of the substance.

The vaporization curve is characterized by the Clausius-Clapeyron equation

Here, the two known points are

(P,T) = (114.9,555)

and

(P,T) = (0.103,114)

We have to find (P,T) = (1,?)

Using the known values in equation,

Using this with P = 1 ATM,

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